UW CHEM 120 Autumn 2012 Name __________________________________________ Section _____________ WS 3: Energy & Nutrition, The Periodic Table, And Atomic Structure 1. Below is the (slightly censored) nutrition label for Fiber One cereal. You may be more familiar with this cereal as the gross one that looks like a bunch of brown worms. a. A box of Fiber Worms, er, Fiber One, says that it contains 16.2 oz (1 lb. 0.2 oz)(455 g) of cereal. How many kilocalories of energy are in one box of this cereal? b. Use the energy value from part a and the table at the end of this worksheet to calculate the following: i. How many hours does a person need to sit to use up the amount of energy contained in a box of Fiber Worms? ii. How many hours could a person run with the same amount of energy? 1 of 4 UW CHEM 120 Autumn 2012 2. Jumbo the elephant is the oldest elephant in the zoo. Poor old Jumbo doesn’t have any teeth, so cannot eat crunchy fruits or grasses like the other elephants. His zookeepers feel guilty about it, but they found out that Fiber Worms soaked in milk become really soft and gross, so they’ve been feeding Jumbo nothing but Fiber Worms and milk for the last year! Jumbo is no doofus, and he has stopped eating food until the zookeepers give him his favorite thing: peanut butter! a. Use the information below (and from problem 1) to find out how many jars of peanut butter the zookeepers will need to feed Jumbo per day. The nutrition label shown below is for natural peanut butter (Jumbo’s favorite). • Peanut butter jar = 16.0 ounces • Jumbo’s daily caloric intake = 150 boxes of fiber worms, plus milk (use the cereal label column that includes ½ cup milk per serving) b. If a box of Fiber Worms costs $3.78, milk costs $5.50 per gallon, and a jar of peanut butter costs $3.99, calculate the daily costs of feeding Jumbo. Which diet is less expensive? (Hint: how much milk will be used per box of cereal?) 2 of 4 UW CHEM 120 Autumn 2012 3. For each of the following elements, give the total number of electrons, the electron configuration, the abbreviated configuration, and the number of valence electrons. # of Abbreviated −
Element Total e Electron Configuration Valence Configuration Electrons H C Ar Ca Br Hg U 4. While doing cutting-­‐edge research, Professor Smith discovers a new element with atomic number 115. She is very excited, and decides to name this new element Dachshundium (Da), after her pet wiener dog. Use the information below to calculate the Atomic Mass of the exciting new element. 284
113
Da
285
113
Da
286
113
Da
Abundance
65.052%
13.069%
21.879%
Mass (amu)
284.003
285.011
286.024
3 of 4 UW CHEM 120 Autumn 2012 Potentially helpful information. Conversion factors: Length 1 m = 1.094 yd 1 mi = 5280 ft 1 ft = 12 in 3
1 km = 10 m 1 mi = 1760 yd 1 Å = 10-­‐10 m 1 m = 102 cm (centimeters) 1 in = 2.54 cm (exact) 1 m = 103 mm (millimeters) Mass 1 kg = 2.205 lb 1 lb = 453.6 g 1 amu = 1.6605 x 10-­‐27 kg 1 dry oz (ounce) = 28.35 g Volume 1 L = 1000 mL = 1.06 qt 1 ft3 = 28.32 L 1 dL = 10-­‐1 L 1 mL = 1 cm3 1 gal = 3.785412 L 1 fluid oz = 29.57 mL Pressure 1 atm = 760. mmHg 1 torr = 1 mmHg 2
1 atm = 14.6959 lb/in (psi) 1 atm = 101,325 Pa 1 Pa = 1 N/m2 1 bar = 100 kPa Energy 1 eV = 1.602 x 10-­‐19 J 1 cal = 4.184 J (exact) 1 kcal = 1 Cal Force 1 N = 0.22481 lb Metric prefixes: tera (T) = 1012 peta (P) = 1015 pico (p) = 10-­‐12 femto (f) = 10-­‐15 Equations: m
ΔT = T final − Tinitial
TF = 1.8 ⋅ TC + 32
q = s ⋅ m ⋅ ΔT
d =
V
TK = TC + 273.15
Physical Data: Ethanol Water 3
3
Density, liquid = 1.000 g/cm Density, liquid = 0.789 g/cm Heat of vaporization = 837.0 J/g Heat of fusion = 107.9 J/g o
Tboil = 78.4 C o
Tfreeze = -­‐114 C Specific heat, liquid = 2.440 J/g.K Specific heat, gas = 1.650 J/g.K 3
Density, solid = 0.917 g/cm Heat of vaporization = 2260 J/g Heat of fusion = 334 J/g o
Tboil = 100 C o
Tfreeze = 0 C Specific heat, solid = 2.11 J/g.K Specific heat, liquid = 4.184 J/g.K Specific heat, gas = 2.080 J/g.K 4 of 4